Chemistry

Chemistry, 25.09.2019 15:00, lekaje2375

#1: which of the following is likely to be a polar compound?
a. chcl3
b. ccl4
c. co2
d. ch4

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Chemistry, 16.07.2019 01:30, mardell2587

This problem has been solved! see the answer suppose a 250. ml flask is filled with 1.0 mol of cl2, 2.0mol of chcl3 and 1.1mol of ccl4 the following reaction becomes possible cl2(g)+chcl3(g) reverseequarrow.gif hcl(g)+ccl4(g) the equilibrium constant k for this reaction 0.933 at the temperature of the flask . calculate the equilibrium molarity of hcl round your answer to two decimal place . . round your answer to two decimal places

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Chemistry, 13.08.2019 05:10, meCreation

The reaction chcl3(g) + cl2(g) ccl4(g) + hcl(g) has been proposed to occur by the following mechanism. cl2 2cl fast equilibrium chcl3 + cl hcl + ccl3 slow step ccl3 + cl ccl4 fast what is the rate law predicted by this mechanism? a. rate = k2[chcl3][cl] b. rate = kexp[chcl3][cl2] c. rate = k2(k1/k-1)1/2[chcl3][cl2]1/2 d. rate = k3[ccl3][cl] e. rate = k1[cl2]

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Chemistry, 12.09.2019 19:20, tishfaco5000

Methane and chlorine react to form four products: ch3cl, ch2cl2, chcl3, and ccl4. at a particular temperature and pressure, 38.4 g of ch4 was allowed to react with excess cl2 and gave 9.2 g ch3cl, 47.1 g ch2cl2, and 109 g chcl3. all the ch4 reacted. (note: the hydrogen that is displaced from the carbon also combines with cl2 to form hcl.)how many grams of ccl4 were formed? how many grams of cl2 reacted with the ch4?

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Chemistry, 18.09.2019 01:00, emalvidrez5205

Consider the following reaction: chcl3(g) + cl2(g) → ccl4(g) + hcl(g) the initial rate of the reaction is measured at several different concentrations of the reactants with the following results: [chcl3] (m) [cl2] (m) initial rate (m/s) 0.010 0.010 0.0035 0.020 0.010 0.0069 0.020 0.020 0.0098 0.040 0.040 0.027 from the data, choose the correct rate law for the reaction. rate=k[chcl3][cl2]2 rate=k[chcl3][cl2]12 rate=k[chcl3]2[cl2] rate=k[chcl3]12[cl2]

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#1: which of the following is likely to be a polar compound?
a. chcl3
b. ccl4
c....

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